Hyperbola

  • Find equation of hyperbola- regular and rectangular, eccentricity, directix, focus, latus rectum, asymptote, relation between a and b.

Hyperbola

Equation of a hyperbola when center is at (h,k) \[\dfrac{(x-h)^{2}}{a^{2}}- \dfrac{(y-k)^{2}}{b^{2}} = 1\] Equation of a hyperbola when center is at (0,0) \[\dfrac{x^{2}}{a^{2}}- \dfrac{y^{2}}{b^{2}} = 1\]

  • Focus of hyperbola: The two points on the transverse axis. These points are what controls the entire shape of the hyperbola since the hyperbola’s graph is made up of all points, P, such that the difference of distances between P and the two foci are equal. To determine the foci you can use the formula: \(a^2 + b^2 = c^2\)

  • Transverse axis: This is the axis on which the two foci are.

  • Asymptotes: The two lines that the hyperbolas come closer and closer to touching. The equation of the asymptotes is always: \[y = \pm \frac{a}{b} x\]

 

Properties of Hyperbola

Solved Examples

Solved Example:

1-6-01

Find the equation of the hyperbola having vertices: ($\pm$5,0), foci: ($\pm$8,0)

Solution:
For hyperbola, $\mathrm{vertices} = (\pm a,0) = (\pm5,0)$,
$\mathrm{foci} = (\pm c,0), c^2 = a^2 + b^2$,
$\mathrm{\ or,\ }b^2 = 8^2 - 5^2 = 64 - 25 = 39$
Equation of a hyperbola, $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 , \dfrac{x^2}{25} - \dfrac{y^2}{39} =1$

Correct Answer: D

Solved Example:

1-6-02

What is the equation of the asymptote of the hyperbola $\dfrac{x^2}{9} - \dfrac{y^2}{4} =1$?

Solution:
$a = 3, b = 2$. Equation of the asymptote is: $y = \pm \dfrac{b}{a}x, y = \pm \dfrac{2}{3}x, 3y = 2x, 2x - 3y = 0$

Correct Answer: A

Solved Example:

1-6-03

Find the equation of the hyperbola whose asymptotes are y = $\pm$ 2x and which passes through (5/2, 3).

Solution:
The equation of the asymptote is, $y = \dfrac{b}{a}x$.
Comparing with the given equation,$\dfrac{b}{a} = 2, b=2a$
So the equation of the hyperbola will become, $\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{(2a)^{2}} = 1$
Substituting the given co-ordinates, \[\dfrac{(5/2)^{2}}{a^{2}}-\dfrac{3^{2}}{(2a)^{2}} = 1, a = 2, b= 4\] \[\dfrac{x^{2}}{2^{2}}-\dfrac{y^{2}}{4^{2}} = 1, 4x^2 - y^2 = 16\]

Correct Answer: D

Solved Example:

1-6-04

4$x^2$ - $y^2$ = 16 is the equation of a/an?

Solution:
\[4x^{2}-y^{2}=16, \dfrac {x^{2}}{4}-\dfrac {y^{2}}{16}=1\] Because of the negative sign of $y^2$ term this is an equation of a hyperbola in the format, $\dfrac {x^{2}}{a^{2}}-\dfrac {y^{2}}{b^{2}}=1$

Correct Answer: B

Solved Example:

1-6-05

Find the eccentricity of the curve: $9x^2 - 4y^2 - 36x + 8y = 4$

Solution:
\begin{align*} 9x^{2}-36x-4y^{2}+8y &=4\\ 9\left( x^{2}-4x+4\right) -36-4\left( y^{2}-2y+1\right) +4 &=4\\ \dfrac {9}{36}\left( x-2\right) ^{2}-\dfrac {4}{36}\left( y-1\right) ^{2} &=\dfrac {36}{36}\\ \dfrac {\left( x-2\right) ^{2}}{4}-\dfrac {\left( y-1\right) ^{2}}{9} &=1 \end{align*} a = 2, b =3 \[b^{2} =a^{2}\left( e^{2}-1\right), 9 =4\left( e^{2}-1\right), 2.25 =e^{2}-1, e =1.80\]

Correct Answer: A

Solved Example:

1-6-06

How far from the x-axis is the focus F of the hyperbola $x^2 - 2y^2 + 4x + 4y + 4 = 0?$

Solution:
\begin{align*} x^{2}-2y^{2}+4x+4y+4 &=0\\ x^{2}+4x+4-2y^{2}+4y&=0\\ \left( x+2\right) ^{2}-2\left( y^{2}-2y\right) &=0\\ \left( x+2\right) ^{2}-2\left( y^{2}-2y+1\right) &=-2\\ \left( x+2\right) ^{2}-2\left( y-1\right) ^{2}&=-2\\ 2\left( y-1\right) ^{2}-\left( x+2\right) ^{2}&=2\\ \left( y-1\right) ^{2}-\dfrac {\left( x+2\right) ^{2}}{2}&=1\\ \end{align*} Comparing with, $\dfrac {Y^{2}}{a^{2}}-\dfrac {X^{2}}{b^{2}}=1, a=1, \ b=\sqrt {2}$ \[X = x+2\ ,\ Y = y-1\] \[e =\dfrac {\sqrt {a^{2}+b^{2}}}{a} =\dfrac {\sqrt {1+2}}{1} =\sqrt {3},\] \[F\equiv \left( 0,\pm ae\right) =\left( 0,\pm \sqrt {3}\right)\] \[X = x+2 \mathrm{\ Focus\ Shifted\ by}\ -2 \mathrm{\ in\ } x \mathrm{\ axis},\] \[Y = y-1 \mathrm{\ Focus\ Shifted\ by}\ 1 \mathrm{\ in\ } y \mathrm{\ axis}\] \[\mathrm{Focus\ Shifted\ by}=\left( -2,1\right)\] \[F_{1} \equiv \left( -2,1+\sqrt {3}\right) =\left( -2,2.73\right),\] \[F_{2}\equiv \left( -2,1-\sqrt {3}\right) =\left( -2,-0.73\right)\]

Correct Answer: C